A puck with mass m = 0.17kg slides along a frictionless horizontal surface at v = 5.3m/s . It hits and sticks to the free end of a spring with k = 17N/m


Find the amplitude and period of the subsequent simple harmonic motion?

Finally was able to solve this myself, but still need help understanding my other question.

To find the amplitude and period of the subsequent simple harmonic motion, we will use the principle of conservation of energy.

First, let's find the initial kinetic energy of the puck before it hits the spring. The kinetic energy of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

KE = (1/2) * 0.17kg * (5.3m/s)^2 = 0.308 kg*m^2/s^2

When the puck hits the spring and sticks to it, the energy will be converted from kinetic energy to potential energy stored in the spring. The potential energy of a spring is given by the equation PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.

Since the puck sticks to the spring, the displacement is the maximum extension of the spring, which is also the amplitude of the subsequent simple harmonic motion. Therefore, we can equate the initial kinetic energy of the puck to the potential energy stored in the spring:

0.308 kg*m^2/s^2 = (1/2) * 17N/m * x^2

Simplifying the equation, we get:

0.308 kg*m^2/s^2 = 8.5 N/m * x^2

x^2 = 0.308 kg*m^2/s^2 / (8.5 N/m) = 0.0362 m^2

Taking the square root of both sides, we find:

x = sqrt(0.0362 m^2) = 0.19 m

Therefore, the amplitude of the subsequent simple harmonic motion is 0.19 m.

Next, let's find the period of the simple harmonic motion. The period of a mass-spring system is given by the equation T = 2π * sqrt(m/k), where m is the mass of the object and k is the spring constant.

T = 2π * sqrt(0.17kg / 17N/m) = 2π * sqrt(0.01 s^2/kg) = 0.63 s

Therefore, the period of the subsequent simple harmonic motion is 0.63 seconds.